## 107.21 Semistable curves

The following lemma will help us understand families of semistable curves.

Lemma 107.21.1. Let $f : X \to S$ be a prestable family of curves of genus $g \geq 1$. Let $s \in S$ be a point of the base scheme. Let $m \geq 2$. The following are equivalent

$X_ s$ does not have a rational tail (Algebraic Curves, Example 53.22.1), and

$f^*f_*\omega _{X/S}^{\otimes m} \to \omega _{X/S}^{\otimes m}$, is surjective over $f^{-1}(U)$ for some $s \in U \subset S$ open.

**Proof.**
Assume (2). Using the material in Section 107.19 we conclude that $\omega _{X_ s}^{\otimes m}$ is globally generated. However, if $C \subset X_ s$ is a rational tail, then $\deg (\omega _{X_ s}|_ C) < 0$ by Algebraic Curves, Lemma 53.22.2 hence $H^0(C, \omega _{X_ s}|_ C) = 0$ by Varieties, Lemma 33.43.12 which contradicts the fact that it is globally generated. This proves (1).

Assume (1). First assume that $g \geq 2$. Assumption (1) implies $\omega _{X_ s}^{\otimes m}$ is globally generated, see Algebraic Curves, Lemma 53.22.6. Moreover, we have

\[ \mathop{\mathrm{Hom}}\nolimits _{\kappa (s)}(H^1(X_ s, \omega _{X_ s}^{\otimes m}), \kappa (s)) = H^0(X_ s, \omega _{X_ s}^{\otimes 1 - m}) \]

by duality, see Algebraic Curves, Lemma 53.4.2. Since $\omega _{X_ s}^{\otimes m}$ is globally generated we find that the restriction to each irreducible component has nonegative degree. Hence the restriction of $\omega _{X_ s}^{\otimes 1 - m}$ to each irreducible component has nonpositive degree. Since $\deg (\omega _{X_ s}^{\otimes 1 - m}) = (1 - m)(2g - 2) < 0$ by Riemann-Roch (Algebraic Curves, Lemma 53.5.2) we conclude that the $H^0$ is zero by Varieties, Lemma 33.43.13. By cohomology and base change we conclude that

\[ E = Rf_*\omega _{X/S}^{\otimes m} \]

is a perfect complex whose formation commutes with arbitrary base change (Derived Categories of Spaces, Lemma 73.25.4). The vanishing proved above tells us that $E \otimes ^\mathbf {L} \kappa (s)$ is equal to $H^0(X_ s, \omega _{X_ s}^{\otimes m})$ placed in degree $0$. After shrinking $S$ we find $E = f_*\omega _{X/S}^{\otimes m}$ is a locally free $\mathcal{O}_ S$-module placed in degree $0$ (and its formation commutes with arbitrary base change as we've already said), see Derived Categories of Spaces, Lemma 73.26.5. The map $f^*f_*\omega _{X/S}^{\otimes m} \to \omega _{X/S}^{\otimes m}$ is surjective after restricting to $X_ s$. Thus it is surjective in an open neighbourhood of $X_ s$. Since $f$ is proper, this open neighbourhood contains $f^{-1}(U)$ for some open neighbourhood $U$ of $s$ in $S$.

Assume (1) and $g = 1$. By Algebraic Curves, Lemma 53.22.6 the assumption (1) means that $\omega _{X_ s}$ is isomorphic to $\mathcal{O}_{X_ s}$. If we can show that after shrinking $S$ the invertible sheaf $\omega _{X/S}$ because trivial, then we are done. We may assume $S$ is affine. After shrinking $S$ further, we can write

\[ Rf_*\mathcal{O}_ X = (\mathcal{O}_ S \xrightarrow {0} \mathcal{O}_ S) \]

sitting in degrees $0$ and $1$ compatibly with further base change, see Lemma 107.9.3. By duality this means that

\[ Rf_*\omega _{X/S} = (\mathcal{O}_ S \xrightarrow {0} \mathcal{O}_ S) \]

sitting in degrees $0$ and $1$, see Duality for Spaces, Lemma 84.3.3^{1}. In particular we obtain an isomorphism $\mathcal{O}_ S \to f_*\omega _{X/S}$ which is compatible with base change since formation of $Rf_*\omega _{X/S}$ is compatible with base change (see reference given above). By adjointness, we get a global section $\sigma \in \Gamma (X, \omega _{X/S})$. The restriction of this section to the fibre $X_ s$ is nonzero (a basis element in fact) and as $\omega _{X_ s}$ is trivial on the fibres, this section is nonwhere zero on $X_ s$. Thus it nowhere zero in an open neighbourhood of $X_ s$. Since $f$ is proper, this open neighbourhood contains $f^{-1}(U)$ for some open neighbourhood $U$ of $s$ in $S$.
$\square$

Motivated by Lemma 107.21.1 we make the following definition.

Definition 107.21.2. Let $f : X \to S$ be a family of curves. We say $f$ is a *semistable family of curves* if

$X \to S$ is a prestable family of curves, and

$X_ s$ has genus $\geq 1$ and does not have a rational tail for all $s \in S$.

In particular, a prestable family of curves of genus $0$ is never semistable. Let $X$ be a proper scheme over a field $k$ with $\dim (X) \leq 1$. Then $X \to \mathop{\mathrm{Spec}}(k)$ is a family of curves and hence we can ask whether or not it is semistable. Unwinding the definitions we see the following are equivalent

$X$ is semistable,

$X$ is prestable, has genus $\geq 1$, and does not have a rational tail,

$X_{\overline{k}}$ is connected, is smooth over $\overline{k}$ apart from a finite number of nodes, has genus $\geq 1$, and has no irreducible component isomorphic to $\mathbf{P}^1_{\overline{k}}$ which meets the rest of $X_{\overline{k}}$ in only one point.

To see the equivalence of (2) and (3) use that $X$ has no rational tails if and only if $X_{\overline{k}}$ has no rational tails by Algebraic Curves, Lemma 53.22.6. This shows that our definition agrees with most definitions one finds in the literature.

Lemma 107.21.3. There exist an open substack $\mathcal{C}\! \mathit{urves}^{semistable} \subset \mathcal{C}\! \mathit{urves}$ such that

given a family of curves $f : X \to S$ the following are equivalent

the classifying morphism $S \to \mathcal{C}\! \mathit{urves}$ factors through $\mathcal{C}\! \mathit{urves}^{semistable}$,

$X \to S$ is a semistable family of curves,

given $X$ a scheme proper over a field $k$ with $\dim (X) \leq 1$ the following are equivalent

the classifying morphism $\mathop{\mathrm{Spec}}(k) \to \mathcal{C}\! \mathit{urves}$ factors through $\mathcal{C}\! \mathit{urves}^{semistable}$,

the singularities of $X$ are at-worst-nodal, $\dim (X) = 1$, $k = H^0(X, \mathcal{O}_ X)$, the genus of $X$ is $\geq 1$, and $X$ has no rational tails,

the singularities of $X$ are at-worst-nodal, $\dim (X) = 1$, $k = H^0(X, \mathcal{O}_ X)$, and $\omega _{X_ s}^{\otimes m}$ is globally generated for $m \geq 2$.

**Proof.**
The equivalence of (2)(b) and (2)(c) is Algebraic Curves, Lemma 53.22.6. In the rest of the proof we will work with (2)(b) in accordance with Definition 107.21.2.

By the discussion in Section 107.6 it suffices to look at families $f : X \to S$ of prestable curves. By Lemma 107.21.1 we obtain the desired openness of the locus in question. Formation of this open commutes with arbitrary base change, because the (non)existence of rational tails is insensitive to ground field extensions by Algebraic Curves, Lemma 53.22.6.
$\square$

Lemma 107.21.4. There is a decomposition into open and closed substacks

\[ \mathcal{C}\! \mathit{urves}^{semistable} = \coprod \nolimits _{g \geq 1} \mathcal{C}\! \mathit{urves}^{semistable}_ g \]

where each $\mathcal{C}\! \mathit{urves}^{semistable}_ g$ is characterized as follows:

given a family of curves $f : X \to S$ the following are equivalent

the classifying morphism $S \to \mathcal{C}\! \mathit{urves}$ factors through $\mathcal{C}\! \mathit{urves}^{semistable}_ g$,

$X \to S$ is a semistable family of curves and $R^1f_*\mathcal{O}_ X$ is a locally free $\mathcal{O}_ S$-module of rank $g$,

given $X$ a scheme proper over a field $k$ with $\dim (X) \leq 1$ the following are equivalent

the classifying morphism $\mathop{\mathrm{Spec}}(k) \to \mathcal{C}\! \mathit{urves}$ factors through $\mathcal{C}\! \mathit{urves}^{semistable}_ g$,

the singularities of $X$ are at-worst-nodal, $\dim (X) = 1$, $k = H^0(X, \mathcal{O}_ X)$, the genus of $X$ is $g$, and $X$ has no rational tail,

the singularities of $X$ are at-worst-nodal, $\dim (X) = 1$, $k = H^0(X, \mathcal{O}_ X)$, the genus of $X$ is $g$, and $\omega _{X_ s}^{\otimes m}$ is globally generated for $m \geq 2$.

**Proof.**
Combine Lemmas 107.21.3 and 107.20.3.
$\square$

Lemma 107.21.5. The morphisms $\mathcal{C}\! \mathit{urves}^{semistable} \to \mathop{\mathrm{Spec}}(\mathbf{Z})$ and $\mathcal{C}\! \mathit{urves}^{semistable}_ g \to \mathop{\mathrm{Spec}}(\mathbf{Z})$ are smooth.

**Proof.**
Since $\mathcal{C}\! \mathit{urves}^{semistable}$ is an open substack of $\mathcal{C}\! \mathit{urves}^{nodal}$ this follows from Lemma 107.18.2.
$\square$

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